Find the maximum value of the function f(x) = x³ – 3x² – 9x + 20.

We are given the function: f(x) = x³ – 3x² – 9x + 20 We need to find its maximum value. Differentiate f(x): f′(x) = 3x² – 6x – 9 Set f′(x) = 0: 3x² – 6x – 9 = 0 Divide all terms by 3: x² – 2x – 3 = 0 Factor: (x – 3)(x + 1) = 0 ⇒ x = 3, –1 Differentiate again: f″(x) = 6x – 6 Now test the sign of f″(x) at critical points:

• At x = –1:
  f″(–1) = 6(–1) – 6 = –12 → Negative ⇒ local maximum
• At x = 3:
  f″(3) = 6(3) – 6 = 12 → Positive ⇒ local minimum

Maximum occurs at x = –1 Substitute into f(x): f(–1) = (–1)³ – 3(–1)² – 9(–1) + 20 = –1 – 3(1) + 9 + 20 = –1 – 3 + 9 + 20 = 25 Therefore, the maximum value of the function f(x) is 25.